46
2 Free Vibration of Single Degree of Freedom System
Fig. 2.25 Damped motion
x 2 = e
−n t 1 exp
−
2π n
p 2 − n 2
C 1 cos
p 2 − n 2 t 1 + C 2 sin
p 2 − n 2 t 1
(2.58)
Dividing Eq. (2.58) by Eq. (2.56), we get
x 1
x 2
= exp
2π n
p 2 − n 2
(2.59)
Taking natural logarithm on both sides gives
ln
x 1
x 2
=
2π n
p 2 − n 2
=
2π
n
p
1 −
n
p
2
(2.60)
If ζ =
n
p
critical damping ratio, then the logarithmic decrement is given by
δ = ln
x 1
x 2
=
2πζ
1 − ζ 2
(2.61)
Usually ζ is a small quantity, so that
1 − ζ 2 ∼ = 1
Then
ln
x 1
x 2
= 2πζ
2 Free Vibration of Single Degree of Freedom System
Fig. 2.25 Damped motion
x 2 = e
−n t 1 exp
−
2π n
p 2 − n 2
C 1 cos
p 2 − n 2 t 1 + C 2 sin
p 2 − n 2 t 1
(2.58)
Dividing Eq. (2.58) by Eq. (2.56), we get
x 1
x 2
= exp
2π n
p 2 − n 2
(2.59)
Taking natural logarithm on both sides gives
ln
x 1
x 2
=
2π n
p 2 − n 2
=
2π
n
p
1 −
n
p
2
(2.60)
If ζ =
n
p
critical damping ratio, then the logarithmic decrement is given by
δ = ln
x 1
x 2
=
2πζ
1 − ζ 2
(2.61)
Usually ζ is a small quantity, so that
1 − ζ 2 ∼ = 1
Then
ln
x 1
x 2
= 2πζ